any solutions?

Abel's formula should make it.

This is an old question. Let $x_{1}=a_{1},x_{i}=a_{i+1}-\sum_{j=1}^{i} a_{j}(2\leq i\leq n)$ , then $x_{i}\geq 0$ , $x_{i+1}-x_{i}=a_{i+1}-2a_{i}.$ So we have $2\sum_{i=1}^{n-1}\frac{a_{i}}{a_{i+1}}=\sum_{i=1}^{n-1}(1-\frac{x_{i+1}-x_{i}}{a_{i+1}})=n-1-\sum_{i=1}^{n-1}\frac{x_{i+1}-x_{i}}{a_{i+1}}=n-\frac{x_{1}}{a_{1}}-\sum_{i=1}^{n-1}\frac{x_{i+1}-x_{i}}{a_{i+1}}=n-\frac{x_{n}}{a_{n}}-\sum_{i=1}^{n-1}x_{i}(\frac{1}{a_{i}}-\frac{1}{a_{i+1}})\leq n.$ The last step is because $a_{i}\leq a_{i+1}$ , which is obvious.